Application of the chain rule

Question

I need to solve the following problem applying the chain rule.

The volume $V$ of a sphere of radius $r$ is given by the formula

$$V=\frac{4\pi r^3}{3}.$$

Suppose that a snowball initially of radius $r=100$ cm is shrinking uniformly at the rate of $1$ cm per hour, so that

$$r=r(t)=100-t.$$

Use the Chain Rule to find a formula for $\frac{dV}{dt}$, the rate at which the volume $V$ of the snowball is shrinking (in cubic centimetres per hour) after $t$ hours have elapsed.

I know that, using the chain rule, $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$ but I can't really understand how to solve this problem.

Answer 1

Yeah, as you know, apply chain rule and replace $r$ by $100-t$.

So, $$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 4\pi(100-t)^2(-1) = -4\pi(100-t)^2$$

As $\frac{dV}{dt} $ is negative, the volume is clearly decreasing.

Answer 2

So You got V=4pir^3/3 r=100-t your task is to derive dV/dt no? the chain rule states that dy/dx=dy/du *du/dx where u is a function in x now as applied to your problem, you find dv/dr first which is 4pir^2{elementary calculus proof} and multiply that by -1 as dr/dt is -1. then chain rule shows that you have found dv/dt. substitue r's equation at the final step and for any time, t, you have the rate of DECREASE in volume of the balloon